Abstract. We illustrate below. You're welcome to make a donation via PayPal. & = n2^{n-1}.\ _\square Calculus Differentiating Exponential Functions From First Principles Key Questions How can I find the derivative of y = ex from first principles? Leaving Cert Maths - Calculus 4 - Differentiation from First Principles Full curriculum of exercises and videos. Conic Sections: Parabola and Focus. Make your first steps in this vast and rich world with some of the most basic differentiation rules, including the Power rule. How do you differentiate f(x)=#1/sqrt(x-4)# using first principles? If it can be shown that the difference simplifies to zero, the task is solved. NOTE: For a straight line: the rate of change of y with respect to x is the same as the gradient of the line. Differentiate from first principles \(y = f(x) = x^3\). Paid link. getting closer and closer to P. We see that the lines from P to each of the Qs get nearer and nearer to becoming a tangent at P as the Qs get nearer to P. The lines through P and Q approach the tangent at P when Q is very close to P. So if we calculate the gradient of one of these lines, and let the point Q approach the point P along the curve, then the gradient of the line should approach the gradient of the tangent at P, and hence the gradient of the curve. \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(a + h) - f(a) }{h} \\ 224 0 obj <>/Filter/FlateDecode/ID[<474B503CD9FE8C48A9ACE05CA21A162D>]/Index[202 43]/Info 201 0 R/Length 103/Prev 127199/Root 203 0 R/Size 245/Type/XRef/W[1 2 1]>>stream First principle of derivatives refers to using algebra to find a general expression for the slope of a curve. By registering you get free access to our website and app (available on desktop AND mobile) which will help you to super-charge your learning process. & = \lim_{h \to 0} \frac{ \sin h}{h} \\ PDF Dn1.1: Differentiation From First Principles - Rmit New Resources. The general notion of rate of change of a quantity \( y \) with respect to \(x\) is the change in \(y\) divided by the change in \(x\), about the point \(a\). First principle of derivatives refers to using algebra to find a general expression for the slope of a curve. Consider a function \(f : [a,b] \rightarrow \mathbb{R}, \) where \( a, b \in \mathbb{R} \). So the coordinates of Q are (x + dx, y + dy). Find the derivative of #cscx# from first principles? Question: Using differentiation from first principles only, determine the derivative of y=3x^(2)+15x-4 tells us if the first derivative is increasing or decreasing. So actually this example was chosen to show that first principle is also used to check the "differentiability" of a such a piecewise function, which is discussed in detail in another wiki. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. Given a function , there are many ways to denote the derivative of with respect to . Click the blue arrow to submit. = &64. & = \lim_{h \to 0} \frac{ (2 + h)^n - (2)^n }{h} \\ & = \lim_{h \to 0^+} \frac{ \sin (0 + h) - (0) }{h} \\ \begin{array}{l l} $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ - Quotient Rule, $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ - Chain Rule, $\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}\arccos(x)=-\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}$, $\frac{d}{dx}\text{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}$, $\frac{d}{dx}\text{arccsc}(x)=-\frac{1}{x\sqrt{x^2-1}}$, Definition of a derivative There is also a table of derivative functions for the trigonometric functions and the square root, logarithm and exponential function. Let's try it out with an easy example; f (x) = x 2. The derivative is a measure of the instantaneous rate of change, which is equal to f' (x) = \lim_ {h \rightarrow 0 } \frac { f (x+h) - f (x) } { h } . Please ensure that your password is at least 8 characters and contains each of the following: You'll be able to enter math problems once our session is over. = & f'(0) \left( 4+2+1+\frac{1}{2} + \frac{1}{4} + \cdots \right) \\ Differentiation from First Principles The First Principles technique is something of a brute-force method for calculating a derivative - the technique explains how the idea of differentiation first came to being. Find the values of the term for f(x+h) and f(x) by identifying x and h. Simplify the expression under the limit and cancel common factors whenever possible. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. So, the answer is that \( f'(0) \) does not exist. \end{array}\]. More than just an online derivative solver, Partial Fraction Decomposition Calculator. Conic Sections: Parabola and Focus. The Derivative Calculator supports solving first, second., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. Note that when x has the value 3, 2x has the value 6, and so this general result agrees with the earlier result when we calculated the gradient at the point P(3, 9). & = \boxed{1}. How do we differentiate from first principles? Knowing these values we can calculate the change in y divided by the change in x and hence the gradient of the line PQ. Basic differentiation rules Learn Proof of the constant derivative rule The equations that will be useful here are: \(\lim_{x \to 0} \frac{\sin x}{x} = 1; and \lim_{x_to 0} \frac{\cos x - 1}{x} = 0\). This is defined to be the gradient of the tangent drawn at that point as shown below. Linear First Order Differential Equations Calculator - Symbolab Log in. First principles is also known as "delta method", since many texts use x (for "change in x) and y (for . What is the second principle of the derivative? \]. As an Amazon Associate I earn from qualifying purchases. Step 4: Click on the "Reset" button to clear the field and enter new values. Check out this video as we use the TI-30XPlus MathPrint calculator to cal. Instead, the derivatives have to be calculated manually step by step. Learn differential calculus for freelimits, continuity, derivatives, and derivative applications. A sketch of part of this graph shown below. Doing this requires using the angle sum formula for sin, as well as trigonometric limits. \]. Everything you need for your studies in one place. Differentiating a linear function We say that the rate of change of y with respect to x is 3. Test your knowledge with gamified quizzes. Forgot password? We can now factor out the \(\sin x\) term: \[\begin{align} f'(x) &= \lim_{h\to 0} \frac{\sin x(\cos h -1) + \sin h\cos x}{h} \\ &= \lim_{h \to 0}(\frac{\sin x (\cos h -1)}{h} + \frac{\sin h \cos x}{h}) \\ &= \lim_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + lim_{h \to 0} \frac{\sin h \cos x}{h} \\ &=(\sin x) \lim_{h \to 0} \frac{\cos h - 1}{h} + (\cos x) \lim_{h \to 0} \frac{\sin h}{h} \end{align} \]. Example : We shall perform the calculation for the curve y = x2 at the point, P, where x = 3. \[\displaystyle f'(1) =\lim_{h \to 0}\frac{f(1+h) - f(1)}{h} = p \ (\text{call it }p).\]. Enter your queries using plain English. \begin{cases} Here are some examples illustrating how to ask for a derivative. We will have a closer look to the step-by-step process below: STEP 1: Let \(y = f(x)\) be a function. w0:i$1*[onu{U 05^Vag2P h9=^os@# NfZe7B Differentiation from First Principles The formal technique for finding the gradient of a tangent is known as Differentiation from First Principles. Write down the formula for finding the derivative from first principles g ( x) = lim h 0 g ( x + h) g ( x) h Substitute into the formula and simplify g ( x) = lim h 0 1 4 1 4 h = lim h 0 0 h = lim h 0 0 = 0 Interpret the answer The gradient of g ( x) is equal to 0 at any point on the graph. How to differentiate 1/x from first principles (limit definition)Music by Adrian von Ziegler Exploring the gradient of a function using a scientific calculator just got easier. # f'(x) = lim_{h to 0} {f(x+h)-f(x)}/{h} #, # f'(x) = lim_{h to 0} {e^(x+h)-e^(x)}/{h} # Because we are considering the graph of y = x2, we know that y + dy = (x + dx)2. Then we have, \[ f\Bigg( x\left(1+\frac{h}{x} \right) \Bigg) = f(x) + f\left( 1+ \frac{h}{x} \right) \implies f(x+h) - f(x) = f\left( 1+ \frac{h}{x} \right). Maxima takes care of actually computing the derivative of the mathematical function. \end{array} # " " = lim_{h to 0} e^x((e^h-1))/{h} # Differentiation from First Principles | TI-30XPlus MathPrint calculator You find some configuration options and a proposed problem below. \(_\square\). ", and the Derivative Calculator will show the result below. Free Step-by-Step First Derivative Calculator (Solver) _.w/bK+~x1ZTtl We can take the gradient of PQ as an approximation to the gradient of the tangent at P, and hence the rate of change of y with respect to x at the point P. The gradient of PQ will be a better approximation if we take Q closer to P. The table below shows the effect of reducing PR successively, and recalculating the gradient. Also, had we known that the function is differentiable, there is in fact no need to evaluate both \( m_+ \) and \( m_-\) because both have to be equal and finite and hence only one should be evaluated, whichever is easier to compute the derivative. I know the derivative of x^3 should be 3x^2 from the power rule however when trying to differentiate using first principles (f'(x)=limh->0 [f(x+h)-f(x)]/h) I ended up with 3x^2+3x. Differentiate from first principles \(f(x) = e^x\). sF1MOgSwEyw1zVt'B0zyn_'sim|U.^LV\#.=F?uS;0iO? Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. This time, the function gets transformed into a form that can be understood by the computer algebra system Maxima. We use this definition to calculate the gradient at any particular point. Differentiation From First Principles - A-Level Revision Joining different pairs of points on a curve produces lines with different gradients. > Differentiation from first principles. Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator. 0 && x = 0 \\ But wait, we actually do not know the differentiability of the function. In doing this, the Derivative Calculator has to respect the order of operations. + #, Differentiating Exponential Functions with Calculators, Differentiating Exponential Functions with Base e, Differentiating Exponential Functions with Other Bases. Get some practice of the same on our free Testbook App. Symbolab is the best derivative calculator, solving first derivatives, second derivatives, higher order derivatives, derivative at a point, partial derivatives, implicit derivatives, derivatives using definition, and more. This expression is the foundation for the rest of differential calculus: every rule, identity, and fact follows from this. [9KP ,KL:]!l`*Xyj`wp]H9D:Z nO V%(DbTe&Q=klyA7y]mjj\-_E]QLkE(mmMn!#zFs:StN4%]]nhM-BR' ~v bnk[a]Rp`$"^&rs9Ozn>/`3s @ The left-hand side of the equation represents \(f'(x), \) and if the right-hand side limit exists, then the left-hand one must also exist and hence we would be able to evaluate \(f'(x) \). Follow the below steps to find the derivative of any function using the first principle: Learnderivatives of cos x,derivatives of sin x,derivatives of xsinxandderivative of 2x, A generalization of the concept of a derivative, in which the ordinary limit is replaced by a one-sided limit. It helps you practice by showing you the full working (step by step differentiation). . We choose a nearby point Q and join P and Q with a straight line. Answer: d dx ex = ex Explanation: We seek: d dx ex Method 1 - Using the limit definition: f '(x) = lim h0 f (x + h) f (x) h We have: f '(x) = lim h0 ex+h ex h = lim h0 exeh ex h We want to measure the rate of change of a function \( y = f(x) \) with respect to its variable \( x \). The derivative is an important tool in calculus that represents an infinitesimal change in a function with respect to one of its variables. & = \lim_{h \to 0} \frac{ 1 + 2h +h^2 - 1 }{h} \\ Using Our Formula to Differentiate a Function. It has reduced by 3. Evaluate the derivative of \(x^2 \) at \( x=1\) using first principle. Pick two points x and x + h. STEP 2: Find \(\Delta y\) and \(\Delta x\). Enter the function you want to differentiate into the Derivative Calculator. It is also known as the delta method. Thermal expansion in insulating solids from first principles This allows for quick feedback while typing by transforming the tree into LaTeX code. \sin x && x> 0. When x changes from 1 to 0, y changes from 1 to 2, and so the gradient = 2 (1) 0 (1) = 3 1 = 3 No matter which pair of points we choose the value of the gradient is always 3. First Principles of Derivatives are useful for finding Derivatives of Algebraic Functions, Derivatives of Trigonometric Functions, Derivatives of Logarithmic Functions. + (4x^3)/(4!) Use parentheses! PDF Differentiation from rst principles - mathcentre.ac.uk You can also choose whether to show the steps and enable expression simplification. Note for second-order derivatives, the notation is often used. They are also useful to find Definite Integral by Parts, Exponential Function, Trigonometric Functions, etc. David Scherfgen 2023 all rights reserved. For example, it is used to find local/global extrema, find inflection points, solve optimization problems and describe the motion of objects. Free derivatives calculator(solver) that gets the detailed solution of the first derivative of a function. Hysteria; All Lights and Lights Out (pdf) Lights Out up to 20x20 Solved Example on One-Sided Derivative: Is the function f(x) = |x + 7| differentiable at x = 7 ? 202 0 obj <> endobj This is also referred to as the derivative of y with respect to x. heyy, new to calc. How do we differentiate a trigonometric function from first principles? \]. \(m_{tangent}=\lim _{h{\rightarrow}0}{y\over{x}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). At a point , the derivative is defined to be . Loading please wait!This will take a few seconds. The third derivative is the rate at which the second derivative is changing. For example, constant factors are pulled out of differentiation operations and sums are split up (sum rule). In each calculation step, one differentiation operation is carried out or rewritten. Create flashcards in notes completely automatically. & = \lim_{h \to 0} \frac{ 2h +h^2 }{h} \\ This limit, if existent, is called the right-hand derivative at \(c\). Did this calculator prove helpful to you? This is called as First Principle in Calculus. Tutorials in differentiating logs and exponentials, sines and cosines, and 3 key rules explained, providing excellent reference material for undergraduate study. So for a given value of \( \delta \) the rate of change from \( c\) to \( c + \delta \) can be given as, \[ m = \frac{ f(c + \delta) - f(c) }{(c+ \delta ) - c }.\]. This means we will start from scratch and use algebra to find a general expression for the slope of a curve, at any value x. Function Commands: * is multiplication oo is \displaystyle \infty pi is \displaystyle \pi x^2 is x 2 sqrt (x) is \displaystyle \sqrt {x} x It is also known as the delta method. both exists and is equal to unity. Plugging \sqrt{x} into the definition of the derivative, we multiply the numerator and denominator by the conjugate of the numerator, \sqrt{x+h}+\sqrt{x}. This hints that there might be some connection with each of the terms in the given equation with \( f'(0).\) Let us consider the limit \( \lim_{h \to 0}\frac{f(nh)}{h} \), where \( n \in \mathbb{R}. Look at the table of values and note that for every unit increase in x we always get an increase of 3 units in y. lim stands for limit and we say that the limit, as x tends to zero, of 2x+dx is 2x. Now, for \( f(0+h) \) where \( h \) is a small negative number, we would use the function defined for \( x < 0 \) since \(h\) is negative and hence the equation. Not what you mean? The x coordinate of Q is x + dx where dx is the symbol we use for a small change, or small increment in x. Example Consider the straight line y = 3x + 2 shown below Then, This is the definition, for any function y = f(x), of the derivative, dy/dx, NOTE: Given y = f(x), its derivative, or rate of change of y with respect to x is defined as. hb```+@(1P,rl @ @1C .pvpk`z02CPcdnV\ D@p;X@U = & 4 f'(0) + 2 f'(0) + f'(0) + \frac{1}{2} f'(0) + \cdots \\ How to differentiate x^3 by first principles : r/maths - Reddit Derivative Calculator - Symbolab Differentiation from First Principles. \[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]. Calculating the gradient between points A & B is not too hard, and if we let h -> 0 we will be calculating the true gradient. Learn about Differentiation and Integration and Derivative of Sin 2x, \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = sinxcosh + cosxsinh sinx\\ = sinx(cosh-1) + cosxsinh\\ {f(x+h) f(x)\over{h}}={ sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinx(cosh-1)\over{h}} + \lim _{h{\rightarrow}0} {cosxsinh\over{h}}\\ = sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} + cosx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = sinx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \times1 = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}) = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}\), Learn about Derivative of Log x and Derivative of Sec Square x, \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = cosxcosh sinxsinh cosx\\ = cosx(cosh-1) sinxsinh\\ {f(x+h) f(x)\over{h}}={ cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosx(cosh-1)\over{h}} \lim _{h{\rightarrow}0} {sinxsinh\over{h}}\\ = cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} sinx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = cosx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \times1 = -sinx\\ f(x)={dy\over{dx}} = {d(cosx)\over{dx}} = -sinx \end{matrix}\), \(\begin{matrix}\ f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = {-2sin({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {-2sin({2x+h\over{2}})sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2cos(x+{h\over{2}}){sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}-2sin(x+{h\over{2}}){sin({h\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}}) = -sinx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = -sinx \end{matrix}\), If f(x) = tanx , find f(x) \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=tanx\\ f(x+h)=tan(x+h)\\ f(x+h)f(x)= tan(x+h) tan(x) = {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\\ {f(x+h) f(x)\over{h}}={ {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosxsin(x+h) sinxcos(x+h)\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {{sin(2x+h)+sinh\over{2}} {sin(2x+h)-sinh\over{2}}\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cosxcos(x+h)}}\\ =1\times{1\over{cosx\times{cosx}}}\\ ={1\over{cos^2x}}\\ ={sec^2x}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = {sec^2x}\\ f(x)={dy\over{dx}} = {d(tanx)\over{dx}} = {sec^2x} \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=sin5x\\ f(x+h)=sin(5x+5h)\\ f(x+h)f(x)= sin(5x+5h) sin(5x) = sin5xcos5h + cos5xsin5h sin5x\\ = sin5x(cos5h-1) + cos5xsin5h\\ {f(x+h) f(x)\over{h}}={ sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sin5x(cos5h-1)\over{h}} + \lim _{h{\rightarrow}0} {cos5xsin5h\over{h}}\\ = sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} + cos5x \lim _{h{\rightarrow}0} {sin5h\over{h}}\\ \text{Put h = 0 in first limit}\\ sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} = sin5x\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} 5\times{{d\over{dh}}sin5h\over{{d\over{dh}}5h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} {5cos5h\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \times5 = 5cos5x \end{matrix}\).

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